1. What pressure would be needed to compress 25.1 mL of hydrogen at 1.01 atm to 25% of its original volume?
2. If the pressure on a 1.04-L sample of gas is doubled at constant temperature, what will be the new volume of the gas?
3. A 1.04-L sample of gas at 759 mm Hg pressure is expanded until its volume is 2.24 L. What will be the pressure in the expanded gas sample (at constant temperature)?
4. What pressure would have to be applied to a 27.2-mL sample of gas at 25 degrees celsius and 1.00 atm to compress its volume to 1.00 mL without a change in temperature?
5. How would you describe absolute zero?
6. A favorite demonstration in introductory chemistry is to illustrate how the volume of a gas is affected by temperature by blowing up a balloon at room temperature and then placing the balloon into a container of dry ice or liquid nitrogen (both of which are very cold). Suppose a balloon containing 1.15 L of air at 25.2 degrees Celsius is placed into a flask containing liquid nitrogen at -78.5 degrees celsius. What will the volume of the sample become (at constant pressure)?
7.Suppose a 1.25 L of argon is cooled from 291 K to 78 K. What will be the new volume of the argon sample?
8. Suppose a gas sample is cooled from 600 K to 300 K. Ho will the new volume of the gas be related to its original volume?
9. Th label of an aerosol spray can contains a warning that the can should not be heated to over 130 degrees F because of the danger of explosion due to the pressure increase as it is heated. Calculate the potential volume of the gas contained in 500-mL aerosol can when it is heated from 25 degrees C to 54 degrees C (approx. 130 degrees F) assuming a constant pressure.
10. A sample of gas has a volume of 127 mL in a boiling water bath at 100 degrees celsius. Calculate the volume of the sample of gas 10 degrees celsius intervals after the heat source is turned off and the gas sample begins to cool down to eh temperature of the laboratory, 20 degrees celsius.
11. At conditions of constant temperature and pressure, the volume of a sample of ideal gas is _____ proportional to the number of moles of gas present.
12. A mathematical expression that summarizes Avogadro's law is ________.
13. If 0.105 mol of helium gas occupies a volume of 2.35 L at a certain temperature and pressure, what volume would 0.337 mol of helium occupy under the same conditions?
14. If 1.00 mol of helium occupies a volume of 22.4 L at 273 K at 1.00 atm, what volume will 1.00 g of helium will occupy under the same conditions?
15. If 3.25 mol of argon gas occupies a volume if 100. L at a particular and temperature and pressure, what volume does 14.15 mol of argon occupy under the same conditions?
16. If 2.71 g of argon gas occupies a volume of 4.21 L, what volume will 1.29 mol of argon occupy under the same conditions?
17. If gaseous mixture is made of 2.41 g of He and 2.79 g of Ne in an evacuated 1.04-L container at 25 degrees celsius, what will be the partial pressure of each gas and the total pressure in the container?
18. A tank contains a mixture of 3.0 mol of N2, 2.0 mol of O2, and 1.0 mol of CO2 at 25 degrees celsius and a total pressure of 10.0 atm. Calculate the partial pressure (in torr) of each gas in the mixture.
19. How many moles of helium gas would be required 2.14-L container to a pressure of 759 mm Hg at 25 degrees celsius? How many moles of neon gas would be required to fill a similar tank to the same pressure at 25 degrees celsius?
20. Calculate what mass of argon gas is required to fill a 20.4-L container to a pressure of 1.09 atm at 25 degrees celsius.
21. What is the pressure in a 245-L tank that contains 5.21 kg of helium at 27 degrees celsius?
22. What mass of helium gas is needed to pressurize a 100.0-L tank to a 255 atm at 25 degrees celsius? What mass of oxygen gas would be needed to pressurize a similar tank to the same specifications?
23. At what temperature will a 1.0 g sample of neon gas exert a pressure of 500. torr in a 5.0 L container?
These are some review questions that can help me study so all the help would be great so I can study from it :) Thank you!
Please help me, I will be grateful!
I agree with Ms. Sue; however, I can give you a couple of tips.
I notice that most of these review questions are for gases with pressure, temperature, moles (or grams) and volume changes.
(P1V1)/T1 = (P2V2)/T2 will work those that don't have grams or moles listed.
For the others, most can be worked with the universal gas law equation, PV = nRT. When using this equation don't forget to change temperature to Kelvin. Kelvin = 273 + degrees celsius.
If you are given grams, you can change to moles by moles = grams/molar mass.
If you run into problems, post A (not dozens) problem and show your work. Please explain what you don't understand about the problem and why you are confused about the next step to take.
i need help
i need help for this quation
1. To find the pressure needed to compress a gas to a certain volume, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
Boyle's Law can be expressed as: P1V1 = P2V2
Given:
V1 = 25.1 mL
P1 = 1.01 atm
V2 = 25% of V1 = (25%)(25.1 mL) = 6.275 mL
Now we can plug these values into the equation and solve for P2:
P1V1 = P2V2
(1.01 atm)(25.1 mL) = P2(6.275 mL)
P2 = (1.01 atm)(25.1 mL) / (6.275 mL)
P2 = 4.06 atm
Therefore, the pressure needed to compress 25.1 mL of hydrogen at 1.01 atm to 25% of its original volume is approximately 4.06 atm.
2. In this question, we have a gas sample with an initial volume (V1) of 1.04 L. If the pressure is doubled and the temperature remains constant, we can use Boyle's Law again to find the new volume (V2).
Boyle's Law: P1V1 = P2V2
Given:
V1 = 1.04 L
P1 = initial pressure
P2 = 2 * P1 (twice the initial pressure)
Rearranging the equation, we get:
V2 = (P1V1) / P2
V2 = (P1 * 1.04 L) / (2 * P1)
V2 = 0.52 L
Therefore, the new volume of the gas sample, when the pressure is doubled at constant temperature, is 0.52 L.
3. Similar to the previous question, we can use Boyle's Law to solve for the pressure in an expanded gas sample.
Given:
V1 = 1.04 L
P1 = 759 mm Hg
V2 = 2.24 L
P2 can be found using the equation: P1V1 = P2V2
P2 = (P1V1) / V2
P2 = (759 mm Hg * 1.04 L) / 2.24 L
P2 ≈ 351 mm Hg
Therefore, the pressure in the expanded gas sample, at constant temperature, is approximately 351 mm Hg.
4. To find the pressure required to compress the gas sample from a larger volume (V1) to a smaller volume (V2) without a change in temperature, we can again use Boyle's Law.
Given:
V1 = 27.2 mL
P1 = 1.00 atm
V2 = 1.00 mL
P2 can be calculated using: P1V1 = P2V2
P2 = (P1V1) / V2
P2 = (1.00 atm * 27.2 mL) / 1.00 mL
P2 = 27.2 atm
Therefore, the pressure required to compress the gas sample to 1.00 mL without a change in temperature is 27.2 atm.
5. Absolute zero is the lowest temperature possible, at which all molecular motion ceases. It is the point at which the particles within a substance have minimal energy and cease to move. In the Celsius and Fahrenheit temperature scales, absolute zero is approximately -273.15 degrees Celsius or -459.67 degrees Fahrenheit.
6. In this scenario, we can use Charles's Law to solve for the new volume of the gas sample when it is subjected to a lower temperature.
Charles's Law states that the volume of a gas is directly proportional to its temperature in Kelvin, at constant pressure.
Given:
V1 = 1.15 L
T1 = 25.2°C = 25.2 + 273.15 K
T2 = -78.5°C = -78.5 + 273.15 K
V2 can be found using the equation: V1 / T1 = V2 / T2
V2 = (V1 * T2) / T1
V2 = (1.15 L * -78.5 + 273.15 K) / (25.2 + 273.15 K)
V2 = 0.743 L
Therefore, the volume of the gas sample will become approximately 0.743 L when placed into the liquid nitrogen at -78.5°C.
7. Similar to the previous question, we can use Charles's Law to calculate the new volume of the gas sample when it is cooled.
Given:
V1 = 1.25 L
T1 = 291 K
T2 = 78 K
V2 can be found using the equation: V1 / T1 = V2 / T2
V2 = (V1 * T2) / T1
V2 = (1.25 L * 78 K) / 291 K
V2 ≈ 0.334 L
Therefore, the new volume of the argon sample, when cooled from 291 K to 78 K, is approximately 0.334 L.
8. When a gas is cooled at a constant pressure, we can use Charles's Law to determine the relationship between the original volume (V1) and the new volume (V2).
Charles's Law states that the volume of a gas is directly proportional to its temperature in Kelvin at constant pressure.
Given:
V1 = original volume
T1 = initial temperature
T2 = final temperature
The relationship is expressed as: V1 / T1 = V2 / T2
If the temperature is halved (T2 = T1/2), the new volume (V2) will also be halved (V2 = V1/2).
Therefore, the new volume of the gas will be half of the original volume when a gas sample is cooled from 600 K to 300 K at constant pressure.
9. To calculate the potential volume of a gas contained in a can when heated, we can use the combined gas law, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law.
Given:
V1 = 500 mL
T1 = 25 °C = 25 + 273.15 K
T2 = 54 °C = 54 + 273.15 K
P1 = constant pressure
Using the combined gas law: (P1 * V1) / T1 = (P2 * V2) / T2
We can rearrange the equation to solve for V2:
V2 = (P2 * V1 * T2) / (P1 * T1)
Substituting the given values, we get:
V2 = (P1 * 500 mL * (54 + 273.15 K)) / (P1 * (25 + 273.15 K))
Simplifying further:
V2 ≈ 729 mL
Therefore, the potential volume of the gas contained in the aerosol can, when heated from 25 °C to 54 °C, is approximately 729 mL assuming a constant pressure.
(To be continued...)